For $n=1$,
$\begin{array}{cl}
& \text{LS} \\
= & \dsum_{k=1}^1 (3k^5+k^3) \\
= & 3(1)^5+(1)^3 \\
= & 4
\end{array}$
$\begin{array}{cl}
& \text{RS} \\
= & \dfrac{1^3(1+1)^3}{2} \\
= & 4
\end{array}$
$\therefore \text{LS}=\text{RS}$.
$\therefore P(1)$ is true.
Assume that $P(m)$ is true for some positive integers $m$.
i.e. $\dsum_{k=1}^m (3k^5+k^3)=\dfrac{m^3(m+1)^3}{2}$.
For $n=m+1$,
$\begin{array}{cl}
& \text{LS} \\
= & \dsum_{k=1}^{m+1}(3k^5+k^3) \\
= & \dsum_{k=1}^m(3k^5+k^3) + 3(m+1)^5+(m+1)^3 \\
= & \dfrac{m^3(m+1)^3}{2}+3(m+1)^5+(m+1)^3 \\
= & \dfrac{(m+1)^3}{2}[m^3+6(m+1)^2+2] \\
= & \dfrac{(m+1)^3}{2}(m^3+6m^2+12m+6+2) \\
= & \dfrac{(m+1)^3}{2}(m^3+6m^2+12m+8) \\
= & \dfrac{(m+1)^3(m+2)^3}{2} \\
= & \text{RS}
\end{array}$
$\therefore P(m+1)$ is also true.
Therefore by mathematical induction, $P(n)$ is true for all positive integers $n$.
